Thank you for your help.

I did the calculations myself and in the last identity in your answer i did not obtain 1/2π multiplier but i obtained 1/N instead. Am I mistaken ?

Thank you for your response. Isn't it z^(n-1) rather than x^(z-1) in the integral ??

That's right sir, thank you. The CTFT case seems to be closed. After the DFT case is closed I'll mark my question answered. What can we say about the DFT case ?

n is a countable number, the sum has countably infinite elements in it. However an integral is performed on real numbers which are uncountably infinite. How can we do that transition from countably infinite numbers to uncountably infinite numbers ? Is it just an approximation ?

letting the "periodicity" of n to go infinity with redefining, say, n = n'/w and letting w to go infinity may change the nature of equation (4). Since int = in't/w after our new definition, letting w go to infinity will cause t/w to converge to 0 (remember, t is finite by definition). This means we do not describe a circular closed contour in (2), Laurent series will not work. If we change the definition of t to eliminate this issue by redefining t within an infinite interval, then we obtain a fraction of infinities t/w = inf/inf. This may yield anything.